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Finding limit

  1. Aug 16, 2016 #1
    1. The problem statement, all variables and given/known data
    limit (5/(2+(9+x)^(0.5))^(cosecx)
    x-->0
    attempt:
    tried applying lim (1+x)^(1/x) = e.
    x->0
    couldn't get anywhere.
     
  2. jcsd
  3. Aug 16, 2016 #2
    Try direct substitution
     
  4. Aug 16, 2016 #3
    wouldn't work.there's a 'cosecx' up there which goes to infinity at x = 0
     
  5. Aug 16, 2016 #4
    Ahhh you're right, sorry
     
  6. Aug 16, 2016 #5
    Is it permissible to use L'hopital's rule?
     
    Last edited: Aug 16, 2016
  7. Aug 16, 2016 #6
    yep.but i would also like to know how to solve it the way i was trying to.i thought abt l'hospital but couldn't figure out how to apply it.
     
  8. Aug 16, 2016 #7
    You have ##\lim_{x\rightarrow 0} ({\frac{5}{2+\sqrt{9+x}}})^{cosec (x)}##

    The key idea here is to take the logarithm. Since the logarithm is a continuous function log of the limit is the limit of the log.

    So the ##e^{\lim_{x\rightarrow 0} \ln({\frac{5}{2+\sqrt{9+x}}})^{cosec (x)}}##

    This reduces to ##e^{\lim_{x\rightarrow 0}\frac{\ln({\frac{5}{2+\sqrt{9+x}})}}{sin (x)}}##

    You can apply l'hopital or use other methods to evaluate the top portion.

    P.S - Next time while asking homework questions, use the template.
     
  9. Aug 16, 2016 #8

    haruspex

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    You can clearly replace the cosec x with 1/x immediately.
    If you then transform the ##\frac 5{2+\sqrt{9+x}}## into the form 1+ some fraction, you can substitute x=0 immediately in the denominator of that fraction, and expand the numerator with the binomial theorem.
     
  10. Aug 31, 2016 #9
    Thanks all for the Help.
     
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